Sum matrix elements group by indices in Python
I have two matrix (same row and column): one with float values, which are grouped by indices in the other matrix. As a result, I want a dictionary or a list with the sums of the elements for each index. Indices always start at 0.
A = np.array([[0.52,0.25,-0.45,0.13],[-0.14,-0.41,0.31,-0.41]])
B = np.array([[1,3,1,2],[3,0,2,2]])
RESULT = {0: -0.41, 1: 0.07, 2: 0.03, 3: 0.11}
I found this solution, but I'm searching for a faster one. I'm working with matrix with 784 x 300 cells and this algorithm takes ~28ms to complete.
import numpy as np
def matrix_sum_by_indices(indices,matrix):
a = np.hstack(indices)
b = np.hstack(matrix)
sidx = a.argsort()
split_idx = np.flatnonzero(np.diff(a[sidx])>0)+1
out = np.split(b[sidx], split_idx)
return [sum(x) for x in out]
If you can help me find a better and plain solution to this problem, I'll be grateful!
EDIT: I made a mistake, time to complete is ~8ms in a 300*10 matrix, but ~28ms in a 784x300.
EDIT2: My A elements are float64, so bincount give me ValueError.
3 answers
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answered 2018-07-10 15:16
Eelco Hoogendoorn
The numpy_indexed package has efficient and simple solutions to this problem (disclaimer: I am its author):
import numpy_indexed as npi keys, values = npi.group_by(B.flatten()).sum(A.flatten()) -
answered 2018-07-10 15:16
user3483203
You can make use of
bincounthere:a = np.array([[0.52,0.25,-0.45,0.13],[-0.14,-0.41,0.31,-0.41]]) b = np.array([[1,3,1,2],[3,0,2,2]]) N = b.max() + 1 id = b + (N*np.arange(b.shape[0]))[:, None] # since you can't apply bincount to a 2D array np.sum(np.bincount(id.ravel(), a.ravel()).reshape(a.shape[0], -1), axis=0)Output:
array([-0.41, 0.07, 0.03, 0.11])As a function:
def using_bincount(indices, matrx): N = indices.max() + 1 id = indices + (N*np.arange(indices.shape[0]))[:, None] # since you can't apply bincount to a 2D array return np.sum(np.bincount(id.ravel(), matrx.ravel()).reshape(matrx.shape[0], -1), axis=0)Timings on this sample:
In [5]: %timeit using_bincount(b, a) 31.1 µs ± 1.74 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each) In [6]: %timeit matrix_sum_by_indices(b, a) 61.3 µs ± 2.62 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each) In [88]: %timeit scipy.ndimage.sum(a, b, index=[0,1,2,3]) 54 µs ± 218 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)(
scipy.ndimage.sumshould be faster on much larger samples) -
answered 2018-07-10 15:40
Thomas Baruchel
The following solution, relying on scipy.ndimage.sum is highly optimized for speed:
import numpy as np A = np.array([[0.52,0.25,-0.45,0.13], [-0.14,-0.41,0.31,-0.41]]) B = np.array([[1,3,1,2], [3,0,2,2]]) import scipy.ndimage print(scipy.ndimage.sum(A, B, index=[0,1,2,3]))You may have to work a little for having the
indexparameter be exactly what you want. It is the list of the indices you want to get in the result. Maybe the following will be a good starting point:print(scipy.ndimage.sum(A,B, index=np.unique(B)))but if you know by advance the list of all indices, it will be more efficient to hard-code it here.
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104 47 78 50
This is all I have for my method so far I need to adjust it to work properly any help is welcomed thanks.
public static void runningSum2DArrayList(ArrayList<ArrayList<Integer>> list, int a) { ArrayList<Integer>li; if (a==3) { //adds up int len=list.size(); int c=0; for (List<Integer> l: list) { int b=0; int j=0; for (int i: l) { li=list.get(len-(b+1)); j=li.get(c)+j; System.out.print(j+ " "); b++; } System.out.println(); c++; } } else if (a==2) { //adds right int j=0; for (List<Integer> l: list) { int b=0; for (int i: l) { if (b==0) { j=l.get(b); } else if (b==1) { j=l.get(b-1)+l.get(b); } else { j=l.get(b)+j; } b++; System.out.print(j+ " "); } System.out.println(); } System.out.println("right"); } else if (a==4) { //adds down int c=0; for (List<Integer> l: list) { int b=0; int j=0; for (int i: l) { li=list.get(b); j=li.get(c)+j; System.out.print(j+ " "); b++; } System.out.println(); c++; } } else if (a==1) { //adds left int j=0; for (List<Integer> l: list) { int len=l.size(); int b=0; for (int i: l) { if (b==0) { j=l.get(len-1); } else if (b==1) { j=l.get(len-1)+l.get(len-2); } else { j=l.get(len-(b+1))+j; } b++; System.out.print(j+ " "); } System.out.println(); } } -
Describing number by sum of numbers
Given the whole number. Describe that by sum of numbers. For example input n=3 output 1+1+1=3 1+2=3
How can i code this?
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Opengl Indexed Rendering with multiple VBOs
I'm a beginner at OpenGL, and I have a problem with rendering objects (I'll call them Entities).
I have a separate Renderer object, which just handles the draw calls, and I pass in the Entity I want to render.
Now the problem I'm having is that when I call
glDrawElements()it doesn't draw anything.I've checked that every array has the correct stuff in it, and all the buffers are valid, also my shader works perfectly with the other rendering method. I've looked at other problems questions here, and I have a working render function with Entities that use normal Vertex Rendering.
My render function
void render(Entity model,EntityShader shader) { glBindVertexArray(model.getVao()); mat4 modelMatrix = mat4(); scale(modelMatrix, vec3(2.0)); translate(modelMatrix, vec3(5.0f,0.0f, 0.0f)); shader.loadTransformationMatrix(modelMatrix); glDrawElements(GL_TRIANGLES,model.getSize(),GL_UNSIGNED_SHORT,(void*)0); glBindVertexArray(0); }My Entity's BufferLoader Part
GLuint vao, iv, pv, nv, tv; glGenBuffers(1, &tv); glGenBuffers(1, &nv); glGenBuffers(1, &pv); glGenBuffers(1, &iv); glGenVertexArrays(1,&vao); glBindVertexArray(vao); glBindBuffer(GL_ARRAY_BUFFER, pv); glBufferData(GL_ARRAY_BUFFER, vertices.size() * sizeof(vec3), &vertices[0], GL_STATIC_DRAW); glVertexAttribPointer(0, 3, GL_FLOAT, GL_FALSE, 3 * sizeof(GLfloat), (GLvoid *)0); glEnableVertexAttribArray(0); glBindBuffer(GL_ARRAY_BUFFER, nv); glBufferData(GL_ARRAY_BUFFER, normals.size() * sizeof(vec3), &normals[0], GL_STATIC_DRAW); glVertexAttribPointer(1, 3, GL_FLOAT, GL_FALSE, 3 * sizeof(GLfloat), (GLvoid *)0); glEnableVertexAttribArray(1); glBindBuffer(GL_ARRAY_BUFFER, tv); glBufferData(GL_ARRAY_BUFFER, uvs.size() * sizeof(vec2), &uvs[0], GL_STATIC_DRAW); glVertexAttribPointer(2, 2, GL_FLOAT, GL_FALSE, 2 * sizeof(GLfloat), (GLvoid *)0); glEnableVertexAttribArray(2); glBindBuffer(GL_ELEMENT_ARRAY_BUFFER, iv); glBufferData(GL_ELEMENT_ARRAY_BUFFER,indices.size()*sizeof(unsigned short), &indices.front(), GL_STATIC_DRAW); glBindVertexArray(0);Here's the data structure:
vector <vec3> vertices; vector <vec2> uvs; vector <vec3> normals; vector <unsigned short>indices;These are my locals at glDrawElments() Thanks to your answers, in advance.
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Sum over some indexes and axis with an irregulat list of list of indices numpy
I want to improve a code where I have to sum over different indices. This is the code:
neighbors = [[1,2,4],[2,0],[1,3],[],[0]] omega = np.random((5,5,3,3)) #Sum over indices 0 (filtered) and 2 suma = np.zeros((5,3)) for j in range(5): for l in range(3): suma[j,l] += np.sum(omega[neighbors[j],j,:,l])Now I improved a little bit my code using the axis parameter of sum:
suma2 = np.zeros((5,3)) for j in range(5): suma2[j,:] += np.sum(omega[neighbors[j],j,:,:],axis=(0,1))I want to know how I can avoid the first loop. I tried creating an array of booleans:
neighbors_bool = np.full((5,5),False,dtype=bool) for j in range(5): neighbors_bool[neighbors[j],j] = TrueBut I don't know how to put it in the sum.
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AMPL how to have a summation constraint
I need a constraint that sums according to the following indices:
subject to shift[1]: x[1,2] + x[1,3] + x[1,4] + x[1,5] + x[1,6] >= 8; subject to shift[2]: x[1,2] + x[2,3] + x[2,4] + x[2,5] + x[2,6] >= 7; subject to shift[3]: x[1,3] + x[2,3] + x[3,4] + x[3,5] + x[3,6] >= 12; subject to shift[4]: x[1,4] + x[2,4] + x[3,4] + x[4,5] + x[4,6] >= 9; subject to shift[5]: x[1,5] + x[2,5] + x[3,5] + x[4,5] + x[5,6] >= 6; subject to shift[6]: x[1,6] + x[2,6] + x[3,6] + x[4,6] + x[5,6] >= 10;What I have is:
param n; # number of shifts possible param demand {i in 1..n}; # demand of workers at each shift var x {1..n, 1..n} >= 0; # number of workers per shift # minimize function subject to shift {t in 1..n}: sum{j in 1..(n)} x[t,j] >= demand[t];This is wrong as it gives the following:
subject to shift[1]: x[1,1] + x[1,2] + x[1,3] + x[1,4] + x[1,5] + x[1,6] >= 8; subject to shift[2]: x[2,1] + x[2,2] + x[2,3] + x[2,4] + x[2,5] + x[2,6] >= 7; subject to shift[3]: x[3,1] + x[3,2] + x[3,3] + x[3,4] + x[3,5] + x[3,6] >= 12; subject to shift[4]: x[4,1] + x[4,2] + x[4,3] + x[4,4] + x[4,5] + x[4,6] >= 9; subject to shift[5]: x[5,1] + x[5,2] + x[5,3] + x[5,4] + x[5,5] + x[5,6] >= 6; subject to shift[6]: x[6,1] + x[6,2] + x[6,3] + x[6,4] + x[6,5] + x[6,6] >= 10;
